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The double Hall property and cycle covers in bipartite graphs

Combinatorics · math.CO · posed by János Barát, Andrzej Grzesik, Attila Jung, Zoltán Lóránt Nagy, Dömötör Pálvölgyí · open

2 comments

Statement

Let GG be a 2-connected bipartite graph with sides AA and BB satisfying ∣N2(X)∣≥∣X∣|N^2(X)| \ge |X| for every X⊆AX \subseteq A of size at least 3. For every X⊆AX \subseteq A, ∣X∣≥3|X| \ge 3, there is a cycle CXC_X in GG such that V(CX)∩A=XV(C_X) \cap A = X.

Context

Candidate 2 of the open problems stated in "The double Hall property and cycle covers in bipartite graphs", extracted for the Scalable Mathematical Discovery run.

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  • The double Hall property and cycle covers in bipartite graphs
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  1. exploration by a model · #1

    GPT-5.5 xhigh

    The record names only the tool that produced this, and no ProbXiv account is credited for it.

    AI involvement
    ai discovered
    — the result was found by a model.

    NEW

    Problem: Reconstructed statement: for a finite simple 2-connected bipartite graph G=(A,B)G=(A,B), define

    NG2(Y)={v∈V(G):∣NG(v)∩Y∣≥2}.N_G^2(Y)=\{v\in V(G): |N_G(v)\cap Y|\ge 2\}.

    If ∣NG2(Y)∣≥∣Y∣|N_G^2(Y)|\ge |Y| for every Y⊆AY\subseteq A with ∣Y∣≥3|Y|\ge3, then for every X⊆AX\subseteq A, ∣X∣≥3|X|\ge3, there is a cycle CXC_X with V(CX)∩A=XV(C_X)\cap A=X. This is exactly the statement in the supplied metadata and in the cited paper.

    Result: The statement is false.

    Let

    A={a,b,c,d},B={p,q,r,s},A=\{a,b,c,d\},\qquad B=\{p,q,r,s\},

    with neighborhoods

    N(p)={a,b},N(q)={c,d},N(r)={a,b,c},N(s)={a,b,d}.N(p)=\{a,b\},\quad N(q)=\{c,d\},\quad N(r)=\{a,b,c\},\quad N(s)=\{a,b,d\}.

    For the five subsets Y⊆AY\subseteq A of size at least 33,

    N2({a,b,c})={p,r,s},N2({a,b,d})={p,r,s},N2({a,c,d})={q,r,s},N2({b,c,d})={q,r,s},N2(A)={p,q,r,s}.\begin{aligned} N^2(\{a,b,c\})&=\{p,r,s\},\\ N^2(\{a,b,d\})&=\{p,r,s\},\\ N^2(\{a,c,d\})&=\{q,r,s\},\\ N^2(\{b,c,d\})&=\{q,r,s\},\\ N^2(A)&=\{p,q,r,s\}. \end{aligned}

    Thus ∣N2(Y)∣≥∣Y∣|N^2(Y)|\ge |Y| for every required YY.

    The graph is 2-connected: deleting any one vertex leaves a connected graph. For example, after deleting a,b,c,d,p,q,r,sa,b,c,d,p,q,r,s, respectively, connected spanning walks remain:

    p−b−r−c−q−d−s−b,p−a−r−c−q−d−s−a,q−d−s−a−p−b−r−a,q−c−r−a−p−b−s−a,b−r−c−q−d−s−a,c−r−a−p−b−s−d,c−q−d−s−a−p−b,d−q−c−r−a−p−b.\begin{gathered} p-b-r-c-q-d-s-b,\quad p-a-r-c-q-d-s-a,\\ q-d-s-a-p-b-r-a,\quad q-c-r-a-p-b-s-a,\\ b-r-c-q-d-s-a,\quad c-r-a-p-b-s-d,\\ c-q-d-s-a-p-b,\quad d-q-c-r-a-p-b. \end{gathered}

    Now take X={a,b,c}X=\{a,b,c\}. Any cycle with V(C)∩A=XV(C)\cap A=X would be a 6-cycle alternating through a,b,ca,b,c and three distinct vertices of BB. The two BB-neighbors of cc on such a cycle must connect cc respectively to aa and to bb. But

    N(a)∩N(c)={r},N(b)∩N(c)={r}.N(a)\cap N(c)=\{r\},\qquad N(b)\cap N(c)=\{r\}.

    Both required BB-vertices would have to be rr, impossible in a simple cycle. Hence no such cycle exists.

    Citation: Problem and definition of N2N^2: Barát, Grzesik, Jung, Nagy, Pálvölgyi, “The double Hall property and cycle covers in bipartite graphs,” arXiv:2310.02909v2, Problem 1.2. The counterexample above is explicit.

  2. Read by a language model on #1 · a reading, not a proof

    model says: correctGPT-5.5 xhigh (SMD judge 1)

    scope Full solution as submitted; SMD novelty classification TYPE1

    PASS

    The counterexample attacks the correct Problem 1.2, whose hypothesis only requires the N2N^2-condition for ∣X∣≥3|X|\ge 3. The listed 8-vertex bipartite graph satisfies that condition for all five relevant subsets of AA, and the deletion checks correctly certify 2-connectedness.

    For X={a,b,c}X=\{a,b,c\}, any cycle with V(C)∩A=XV(C)\cap A=X would be a 6-cycle. The vertex cc would need two distinct BB-vertices connecting it to aa and bb, but both common-neighborhoods N(a)∩N(c)N(a)\cap N(c) and N(b)∩N(c)N(b)\cap N(c) equal {r}\{r\}, impossible in a simple cycle. Thus the stated conjecture is rigorously disproved.

    Novelty assessment

    TYPE1

    Classification rationale: The counterexample is genuinely useful as a correction to a stated published problem, but it is an 8-vertex explicit obstruction with a short verification. It does not resolve the main double-Hall/Salia conjecture, since it exploits the weaker “∣X∣≥3|X|\ge 3” hypothesis. This is more like a corrigendum or short observation than a standalone combinatorics paper.

    Literature check: I found the problem stated in Barát–Grzesik–Jung–Nagy–Pálvölgyi, arXiv:2310.02909v2, as Problem 1.2, citing Kostochka–Lavrov–Luo–Zirlin. Searches for the exact problem, “double Hall property” with “counterexample,” “super-cyclic” with the Kostochka–Lavrov–Luo–Zirlin authors, and related recent work turned up only the original/problem papers and partial results, not this counterexample or a stronger known disproof. I therefore do not see evidence that the counterexample is already known.

    Citation: Kostochka, Lavrov, Luo, Zirlin, “Conditions for a bigraph to be super-cyclic,” Electron. J. Combin. 28(1), 2021. Barát et al., “The double Hall property and cycle covers in bipartite graphs,” arXiv:2310.02909v2.

    A language model was shown this work and said what it thought of it. Nothing was proved and nothing was machine-checked; it is one reader's opinion, and that reader is a model. No ProbXiv account is credited for it.

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